639. DEF. Let f (x) be some single-valued algebraical function of x, not containing x as an index, such that for any complex values of x and y, then, if f(x) xf (y)=f(x+y), f(x) = X and f(1) = a, X is said to be a value of the xth power of a; and x is said to be a logarithm of X to base a. Thus the several forms of ƒ (x) which will satisfy the conditions that f(1) = a, and f(x) xf(y) = f (x + y), will be the several values of a". And the several values of x, which will satisfy the equation f(x) = X for any of the above forms of f, will be the values of loga for that particular form of f. 640. To find the forms of f (x). Suppose that f(x) = c + cx + С2x2 + €3ï3 + ... Since the relation f(x) ×f(y) = f(x + y) requires that f(0) = 1, we see that Co = 1. The same relation gives = 1 + c (x + y) + c2 (x + y)2 + + Cr (x + y)" + ... ... = (1 + cx + c2x2 + + Cr-12-1+...) (1+cy + ...). By multiplying out the term c, (x + y)", the coefficient of y is clearly rc,-. Hence equating the coefficients of yx2 ̄1, Putting successively equal to 1, 2, 3,... we have The coefficient c is obtained from the condition that ƒ(1) shall have any given value a; i.e. c is any solution of the equation If c is any root of this equation, the several values of a are the several corresponding values of 641. To find the nature of the roots of the equation f(x) = 1, I. f(x) = 1 has no real root, except x=0. For, if x is real, f(x) is the real positive value of {ƒ(1)}. But ƒ(1)>1. Hence all real positive powers of f(1) are >1; and all real negative powers of f(1) are ◄ < 1. Hence f(x) cannot for value of x, other than zero. = any real II. ƒ(x) = 1 has no complex root. For, since the coefficients in f(x) are real, if a + bi were a root of f(x) = 1, a – bi would be a root. If then both f(a + bi) and f(abi) were equal to 1, f(2a) would be equal to 1; .. by I., a must be zero. Hence all the roots of ƒ (x) = 1 are wholly imaginary. III. Any integral multiple of the common measure of any two roots of ƒ (x) = 1 must also be a root. Let rui and r'ui be two roots having the common measure μi, so that r and r' are integers. == Then, since f(x) ׃ (y) =ƒ (x + y), and since ƒ(rμi) = 1 and f(r'ui) 1, therefore = f(pr~p'r') μi = 1, where p and p' are any integers. But, by the theory of numbers, we may make pr~p'r' = 1. Hence μi is a root of f(x) = 1; and .. any multiple of μi is a root. IV. All the roots of f(x) = 1 must be integral multiples of some common root. For, if μi and any root outside the series rui have any common measure, all the multiples of this common measure will be roots. And these will include the series rμi. Hence, as long as we take two commensurable roots, all the roots will be included in the multiples of their common measure. But, if two roots λi and A'i were incommensurable, we could λ find a fraction as near as we please to so that pλ'i - qλi, Χ which would also be a root, could be made as small as we please. And since any multiple of this root would be a root, it would follow that any wholly imaginary value whatever would be a root. That is, writing x = iy, the equations would be true for all real values whatever. But this is impossible since the coefficients of these equations are not zero. Hence The roots of f (x) = 1 are all the positive and negative integral multiples of some wholly imaginary root which we may call 2πi.* * The introduction of the symbol will be justified later by showing that it has here its usual value. 642. To find the general form of the values of log ab. The general values of log b are the roots of Let a be any one of the roots of this equation: and λa + k any other root. Then, the left hand being ƒ (c), we have .. k is any solution of the equation ƒ (x) = 1; i.e. k = 2nπi. Particular cases of this result should be carefully noted. + = =t. Then for a write e. Then one value of λe is 1. Thus the general values of log are λό + 2ηπί 1 + 2nπi Or again, put b = 1. Then one value of λb is 0. Thus Putting m and n equal to 0, we see that λb is a special value of log.b. Putting n = 0, we see that 2mi is a value of log.l. And in general we find: Each logarithm, such as logab, has a double infinity of values, which is obtained by giving a single infinity of values to log b and to logea. 643. To find the general values of a*. may be called the expansional value of e*. Thus, since the general value of ao is The general values of a are the expansional values of ea+2n-1), for all integral values of n. Assuming then that a may be complex, we will apply this to the three cases according as x is wholly real, wholly imaginary, or complex. (1) Let x be wholly real = p/q say. Then the expansional values of e (λea+2nxi)p/q are those of |